BeautifulSoup takes forever, can this be done faster?

Question

I'm using a Raspberry Pi 1B+ w/ Debian Linux:

Linux rbian 3.18.0-trunk-rpi #1 PREEMPT Debian 3.18.5-1~exp1+rpi16 (2015-03-28) armv6l GNU/Linux

As part of a larger Python program I'm using this code:

#!/usr/bin/env python

import time
from urllib2 import Request, urlopen
from bs4 import BeautifulSoup

_url="http://xml.buienradar.nl/"

s1  = time.time()
req = Request(_url)
print "Request         = {0}".format(time.time() - s1)
s2 = time.time()
response = urlopen(req)
print "URLopen         = {0}".format(time.time() - s2)
s3 = time.time()
output = response.read()
print "Read            = {0}".format(time.time() - s3)
s4 = time.time()
soup = BeautifulSoup(output)
print "Soup (1)        = {0}".format(time.time() - s4)

s5 = time.time()
MSwind = str(soup.buienradarnl.weergegevens.actueel_weer.weerstations.find(id=6350).windsnelheidms)
GRwind = str(soup.buienradarnl.weergegevens.actueel_weer.weerstations.find(id=6350).windrichtinggr)
ms = MSwind.replace("<"," ").replace(">"," ").split()[1]
gr = GRwind.replace("<"," ").replace(">"," ").split()[1]
print "Extracting info = {0}".format(time.time() - s5)

s6 = time.time()
soup = BeautifulSoup(urlopen(_url))
print "Soup (2)         = {0}".format(time.time() - s6)

s5 = time.time()
MSwind = str(soup.buienradarnl.weergegevens.actueel_weer.weerstations.find(id=6350).windsnelheidms)
GRwind = str(soup.buienradarnl.weergegevens.actueel_weer.weerstations.find(id=6350).windrichtinggr)
ms = MSwind.replace("<"," ").replace(">"," ").split()[1]
gr = GRwind.replace("<"," ").replace(">"," ").split()[1]
print "Extracting info = {0}".format(time.time() - s5)

When I run it, I get this output:

Request         = 0.00394511222839
URLopen         = 0.0579500198364
Read            = 0.0346400737762
Soup (1)        = 23.6777830124
Extracting info = 0.183892965317
Soup (2)         = 36.6107468605
Extracting info = 0.382317781448

So, the BeautifulSoup command takes about half a minute to process the _url. I would really love it if this could be done in under 10 seconds.

Any suggestions that would significantly speed up the code (by at least -60%) would be extremely welcome.

Answer 1

Using requests and regular expressions can be a lot shorter and faster. For such relatively simple data gathering regexes work fine.

#!/usr/bin/env python

from __future__ import print_function
import re
import requests
import time

_url = "http://xml.buienradar.nl/"
_regex = '<weerstation id="6391">.*?'\
         '<windsnelheidMS>(.*?)</windsnelheidMS>.*?'\
         '<windrichtingGR>(.*?)</windrichtingGR>'

s1 = time.time()
br = requests.get(_url)
print("Request         = {0}".format(time.time() - s1))
s5 = time.time()
MSwind, GRwind = re.findall(_regex, br.text)[0]
print("Extracting info = {0}".format(time.time() - s5))
print('wind speed', MSwind, 'm/s')
print('wind direction', GRwind, 'degrees')

On my desktop (which is not a raspberry, though :-) ) this runs very fast;

Request         = 0.0723416805267334
Extracting info = 0.0009412765502929688
wind speed 2.35 m/s
wind direction 232.6 degrees

Of course this particular regex would fail if the windsnelheidMS and windrichtingGR tags were reversed. But given that the XML is most probably computer-generated that doesn't seem likely. And there is an solution for it. By first using a regex to capture the text between <weerstation id="6391"> and </weerstation>, and then use two other regexes to find the wind speed and direction.

Answer 2

Install the lxml library; once installed BeautifulSoup will use it as the default parser.

lxml parser the page using the libxml2 C library, which is significantly faster than the default html.parser backend, implemented in pure Python.

You can then also parse the page as XML instead of as HTML:

soup = BeautifulSoup(output, 'xml')

Parsing your given page with lxml should be faster; I can parse the page almost 50 times per second:

>>> timeit("BeautifulSoup(output, 'xml')", 'from __main__ import BeautifulSoup, output', number=50)
1.1700470447540283

Still, I wonder if you are missing some other Python acceleration libraries, as I certainly cannot reproduce your results even with the built-in parser:

>>> timeit("BeautifulSoup(output, 'html.parser')", 'from __main__ import BeautifulSoup, output', number=50)
1.7218239307403564

Perhaps you are memory constrained and the large-ish document causes your OS to swap memory a lot? Memory swapping (writing pages to disk and loading other pages from disk) can bring even the fastest programs to a grinding halt.

Note that instead of using str() on tag elements and splitting off the tags, you can get the value from a tag simply by using the .string attribute:

station_6350 = soup.buienradarnl.weergegevens.actueel_weer.weerstations.find(id=6350)
ml = station_6350.windsnelheidMS.string
gr = station_6350.windrichtingGR.string

If you are using the XML parser, take into account that tagnames must match case (HTML is a case-insensitive mark-up language).

Since this is an XML document, another option would be to use the lxml ElementTree model; you can use XPath expressions to extract the data:

from lxml import etree 

response = urlopen(_url)
for event, elem in etree.iterparse(response, tag='weerstation'):
    if elem.get('id') == '6350':
        ml = elem.find('windsnelheidMS').text
        gr = elem.find('windrichtingGR').text
        break
    # clear elements we are not interested in, adapted from
    # http://stackoverflow.com/questions/12160418/why-is-lxml-etree-iterparse-eating-up-all-my-memory
    elem.clear()
    for ancestor in elem.xpath('ancestor-or-self::*'):
        while ancestor.getprevious() is not None:
            del ancestor.getparent()[0]

This should only build the minimal object tree required, clearing out the weather stations you don't need as you go along the document.

Demo:

>>> from lxml import etree
>>> from urllib2 import urlopen
>>> _url = "http://xml.buienradar.nl/"
>>> response = urlopen(_url)
>>> for event, elem in etree.iterparse(response, tag='weerstation'):
...     if elem.get('id') == '6350':
...         ml = elem.find('windsnelheidMS').text
...         gr = elem.find('windrichtingGR').text
...         break
...     # clear elements we are not interested in
...     elem.clear()
...     for ancestor in elem.xpath('ancestor-or-self::*'):
...         while ancestor.getprevious() is not None:
...             del ancestor.getparent()[0]
... 
>>> ml
'4.64'
>>> gr
'337.8'